# What is the derivative of 1/x using the quotient rule?

## Question:

What is the derivative of {eq}\frac{1}{x} {/eq} using the quotient rule?

## Quotient Rule

The quotient rule is the method to determine the derivative of a function which is the ratio of two differentiable functions.

{eq}\dfrac{d}{{dy}}\left[ {\dfrac{{f\left( y \right)}}{{g\left( y \right)}}} \right] = \dfrac{{\dfrac{d}{{dy}}\left[ {f\left( y \right)} \right]g\left( y \right) - f\left( y \right)\dfrac{d}{{dy}}\left[ {g\left( y \right)} \right]}}{{{{\left[ {g\left( y \right)} \right]}^2}}}{/eq}

We take the derivative of f(y) multiplied by g(y) and subtract f(y) multiplied by the derivative of g(y) and divide the whole expression by {eq}{\left[ {g\left( y \right)} \right]^2} .{/eq}

If the two functions, f(x) and g(x) are differentiable then, according to the quotient rule,

{eq}{\left( {\dfrac{f}{g}} \right)^\prime } = \dfrac{{f'g - fg'}}{{{g^2}}}{/eq}

So, the derivation of the function {eq}\dfrac{1}{x}{/eq} using the quotient rule is given as,

{eq}\begin{align*} \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) &= \dfrac{{x\dfrac{d}{{dx}}\left( 1 \right) - 1\dfrac{d}{{dx}}\left( x \right)}}{{{x^2}}}\\ \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) &= \dfrac{{\left( {x \times 0} \right) - \left( {1 \times 1} \right)}}{{{x^2}}}\\ &= \dfrac{{ - 1}}{{{x^2}}} \end{align*}{/eq} 